91Ó°ÊÓ

We are going to sell cola at a football game. The demand for the cola at the game is denoted by X and follows a continuous distribution with a probability density function\(f\left( x \right)\).

The profit is g cents for each liter. The loss is c cents when we order cola but cannot sell it.

Suppose we order P liters of cola and the demand is x. Now when\(\left( {P \le x} \right)\), that is, the quantity of the order is less or equal to the demand, we make the profit of g cents.

If \(\left( {P > x} \right)\),that is the quantity of the order is greater than demand then we make the net gain of \(\left( {gx - c\left( {P - x} \right)} \right)\) cents.

Now the expected gain, G is given by,

\(\begin{aligned}{c}E\left( G \right) &= \int\limits_P^\infty {\left( {gP} \right)f\left( x \right)dx + \int\limits_0^P {\left( {gx - c\left( {P - x} \right)} \right)f\left( x \right)dx} } \\ &= gP\left( {1 - \int\limits_0^P {f\left( x \right)dx} } \right) + \int\limits_0^P {\left( {gx - cP + cx} \right)f\left( x \right)dx} \\ &= gP\left( {1 - F\left( P \right)} \right) + \left( {g + c} \right)\int\limits_0^P {xf\left( x \right)dx - cP\int\limits_0^P {f\left( x \right)dx} } \\ &= gP\left( {1 - F\left( P \right)} \right) + \left( {g + c} \right)\int\limits_0^P {xf\left( x \right)dx - cPF\left( P \right)} \end{aligned}\)

Now,

\(\begin{aligned}{c}\left( {g + c} \right)\int\limits_0^P {xf\left( x \right)dx = \left( {g + c} \right)\left( {\left\{ {x\int\limits_0^P {f\left( x \right)dx} } \right\} - \int\limits_0^P {\left\{ {\frac{\partial }{{\partial x}}x\int\limits_0^P {f\left( x \right)dx} } \right\}} } \right)} \\ = \left( {g + c} \right)\left( {\left\{ {xF\left( x \right)} \right\}_0^P - \int\limits_0^P {\left\{ {F\left( P \right)} \right\}dx} } \right)\\ = \left( {g + c} \right)\left( {\left\{ {PF\left( P \right)} \right\} - PF\left( P \right)} \right)\\ = 0\end{aligned}\)

Thus, the expected gain is,

\(E\left( G \right) = gP\left( {1 - F\left( P \right)} \right) - cPF\left( P \right) \cdots \left( 1 \right)\)

To maximize the expected gain we have to find the value of P. We have to differentiate E(G) with respect to P. So,

\(\begin{aligned}{c}\frac{{\partial E\left( G \right)}}{{\partial P}} &= \frac{\partial }{{\partial P}}\left( {gP\left\{ {1 - F\left( P \right)} \right\} - cPF\left( P \right)} \right)\\ &= \frac{\partial }{{\partial P}}\left( {gP - gPF\left( P \right) - cPF\left( P \right)} \right)\\ &= \frac{\partial }{{\partial P}}\left( {gP - \left( {g + c} \right)PF\left( P \right)} \right)\end{aligned}\)

By using the product rule, we get,

\(\frac{{\partial E\left( G \right)}}{{\partial P}} = g - \left( {g + c} \right)F\left( P \right)\)

Now, by equating the differential with respect to 0, we get,

\(\begin{aligned}{c}\frac{{\partial E\left( G \right)}}{{\partial P}} = 0\\ \Rightarrow g - \left( {g + c} \right)F\left( P \right) = 0\\ \Rightarrow F\left( P \right) = \frac{g}{{\left( {g + c} \right)}} \cdots \left( 2 \right)\end{aligned}\)

Thus, we should order that amount of cola which amount satisfies the equation (1).